# Proving the existence of non-minimum critical points

Let $$f:\mathbb{R}^n \rightarrow \mathbb{R}$$ be an analytic function such that $$\lim_{|\vec{x}|\rightarrow \infty}f(\vec{x}) = \infty$$ Let $$M$$ be the set of points at which $$f$$ achieves its minimum value. Assume $$M$$ has more than one element.

1)Assume $$M$$ is a discrete set. Prove there exists a critical point of $$f$$ which is not in $$M$$ or construct a counterexample.

2)Is this still valid if we drop the assumption of $$M$$ being discrete?

• For (1) and $n=1$ you can use Rolle's theorem between two consecutive elements of $M$. – conditionalMethod Nov 17 '19 at 13:39

This is an immediate consequence of the mountain pass lemma in the calculus of variations, detailed e.g. in Evans' book [1] on page 480, and thus holds in very general (possibly infinite dimensional) spaces. However, the proof of that theorem is kind of tricky and there's a number of possible pitfalls on the way.

In our case, I think we can slightly simplify the proof as follows. Let $$x,y$$ be two points in $$M$$, and let $$k = \inf_\gamma\max_{t\in[0,1]}f\circ \gamma(t)$$, where $$\gamma$$ is a $$C^1$$-curve from $$x$$ to $$y$$. By discreteness of $$M$$, there exists a circle $$C_\epsilon(x)$$ around $$x$$, not containing $$y$$, with $$\min_{\omega \in C_\epsilon(x)}f(\omega) > \min f =: m$$. Since $$\gamma$$ has to cross $$C_\epsilon(x)$$, we infer $$k>m$$.

The critical points $$\{x \in \mathbb{R}^n: \nabla f(x) = 0\}$$ of $$f$$ form a closed set, since $$\nabla f$$ is continuous, and from $$\lim_{|x|\rightarrow \infty}f(x) = \infty$$ we conclude that $$f^{-1}([a,b])$$ is a compact set for all $$a,b\in \mathbb{R}$$. In combination this implies that the critical values of $$f$$ form a closed set as well. Supposing that $$k$$ is not a critical value, there exists $$\delta >0$$ such that $$[k-\delta,k+\delta]$$ contains no critical values. The set $$A := f^{-1}([k-\delta,k+\delta])$$ is a compact set again, hence there exists $$\kappa > 0$$ such that $$|\nabla f(x)| \geq \kappa$$ on $$A$$. Considering the gradient flow $$\Phi(x,t)$$ given by $$\partial_t \Phi(x,t) = -(\nabla f)\circ\Phi(x,t)$$, the above estimate yields $$f \circ \Phi(x,\frac{2\delta}{\kappa^2})\leq k-\delta$$ for all $$x\in A$$.

Now, let $$\gamma$$ be a curve such that $$\max_{t\in[0,1]}f\circ \gamma(t) < k + \delta$$. Then $$\gamma_1(\tau) := \Phi(\gamma(\tau), \frac{2\delta}{\kappa^2})$$ defines a curve with $$\max_{t\in[0,1]}f\circ \gamma_1(t) \leq k - \delta$$, which still connects $$x$$ and $$y$$, contradicting the definition of $$k$$.

The second claim is certainly wrong for smooth functions, consider for example the smooth function given by $$f(x) = x e^{-\frac{1}{x^2-1}}$$ for $$|x|>1$$ and $$f(x) = 0$$ for $$x \in [-1,1]$$. However, I have a feeling that it might actually still be true for analytic functions satisfying $$\lim_{|x|\rightarrow \infty}f(x) = \infty$$ and $$|M|\geq 2$$, but for deeper reasons than the above.

[1] Evans, Lawrence C., Partial differential equations, Graduate Studies in Mathematics. 19. Providence, RI: American Mathematical Society (AMS). xvii, 662 p. (1998). ZBL0902.35002.